No because 3 is a whole number and 1/8 is not
To see how they are related, let us solve the two equations.
FIRST EQUATION
![m + 5 = 72](https://tex.z-dn.net/?f=m%20%2B%205%20%3D%2072)
![\rightarrow \: m = 72 - 5](https://tex.z-dn.net/?f=%20%5Crightarrow%20%5C%3A%20m%20%3D%2072%20-%205)
![\rightarrow \: m = 67](https://tex.z-dn.net/?f=%20%5Crightarrow%20%5C%3A%20m%20%3D%2067)
SECOND EQUATION
![m + 72 = 5](https://tex.z-dn.net/?f=m%20%2B%2072%20%3D%205)
![\rightarrow \: m = 5 - 72](https://tex.z-dn.net/?f=%20%5Crightarrow%20%5C%3A%20m%20%3D%205%20-%2072)
![\rightarrow \: m = - 67](https://tex.z-dn.net/?f=%20%5Crightarrow%20%5C%3A%20m%20%3D%20-%2067)
Hence the solutions to the two equations are ADDITIVE INVERSE of each other.
Answer: Choice 2) A, B, C
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Explanation:
Use a calculator to find that,
- 946/32 = 29.5625
- 1157/11 = 105.1818 approximately
- 3524/44 = 80.0909 approximately
- 37392/82 = 456
The first three results have a decimal portion as the answer. So there is a remainder here. The remainder is the leftover bit that couldn't make another full value. Only choice D results in a whole number that isn't a decimal value, so this does not have a remainder.
As a smaller example, let's say we had 20 cookies and 3 people. If we want to divide the cookies evenly, then each person gets 20/3 = 6 full cookies and there would be 2 cookies left over (6*3 = 18 are eaten so 20-18 = 2 is left over). Notice that 20/3 = 6.67 approximately. The non-whole number decimal value indicates we have a remainder. If we had 21 cookies, then each person gets 21/3 = 7 full cookies with nothing left over, so there's no remainder here.
The number of possibilities that are there to choose 3 balls is 6.
<h3>How to illustrate the information?</h3>
From the information, we have a bottomless bag of 3 different kinds of balls: blue, green, and red.
The number of possibilities that are there to choose 3 balls will be:
= 3!
= 3 × 2 × 1
= 6 possibilities.
Therefore, the number of possibilities that are there to choose 3 balls is 6.
Learn more about permutations on:
brainly.com/question/4658834
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